18. SHORT GAPS BETWEEN PRIMES.
In this section we shall prove that the gap between two consecutive primes of size x can
be much smaller than the average, log x:
Our goal is to show that if p1 = 2 < p2 = 3 < . . . is the sequence of prime numbers
then
pn+1 − pn
lim inf
= 0.
n→∞
log pn
Researchers had had little success in showing that there are gaps between consecutive
primes that are significantly smaller than the average. However a simple recent method of
Goldston, Pintz and Yildirim counts primes in short intervals with simple weighting functions, which themselves can be easily estimated using the Bombieri-Vinogradov theorem,
and implies the above. Using deeper techniques the method can be used to prove that
there are infinitely many pairs of consecutive primes for which
pn+1 − pn ¿ (log pn )1/2+² .
The Elliott-Halberstam conjecture states that, for any A > 0, we have
¯
¯
X
¯
x ¯¯
x
¯
max ¯θ(x; q, a) −
(18.1)
¿A
¯
φ(q)
(log x)A
(a,q)=1
q≤Q
for any Q ≤ x1−² . We saw in chapter 13 that this holds for Q ≤ x1/2 /(log x)B by the
Bombieri-Vinogradov theorem. We warm up for our main result with another striking
result of Goldston, Pintz and Yildirim:
Theorem 18.1. Suppose that (18.1) holds for Q = xθ , for some fixed θ > 20
21 . Then there
are infinitely many pairs of consecutive primes pn < pn+1 such that pn+1 − pn ≤ 20.
One can improve the 20 here to 16 using other methods. If (18.1) holds for Q = xθ , for
any fixed θ > 12 then there are infinitely many pairs of consecutive primes pn < pn+1 for
which pn+1 − pn ¿θ 1.
Actually we prove somewhat more than Theorem 18.1: Let H be a set of linear forms
Qk
{ai x + bi : i = 1, 2, . . . , k} where bi aj 6= ai bj if i 6= j; and define PH (n) := i=i (ai n + bi ).
Let νH (p) denote the number of distinct residue classes n (mod p) for which p divides
PH (n), and extend this definition to νH (d) for squarefree integers d by multiplicativity.
Define the singular series
¶−k µ
¶
Yµ
1
νH (p)
S(H) =
1−
1−
p
p
p
If S(H) 6= 0 then H is called admissible. Thus H is admissible if and only if νH (p) < p for
all p.
Typeset by AMS-TEX
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MAT6684
Proposition 18.2. Suppose that (18.1) holds for Q = xθ , for some fixed θ > 20
21 . Then
for any
¯ admissible 7-tuple H, there are infinitely many n such that at least two members
of H¯x=n are prime.
To deduce Theorem 18.1 from Proposition 18.2 note that
{x − 10, x − 8, x − 4, x − 2, x + 2, x + 8, x + 10}
is an admissible 7-tuple with each ai = 1 (and |b7 − b1 | as small as possible).
For any non-zero integer b whose prime factors are all ≥ 11, the set {210ix + b : 1 ∈ I}
is admissible for any set, I, of seven distinct integers. Taking I = {1, 2, 3, 4, 5, 6, 7} we
deduce that there exists r/s > 1 with 1 ≤ s < r ≤ 7 such that there are infinitely many
pairs of primes p, q for which (p − b)/(q − b) = r/s. Taking I = {mi : 0 ≤ i ≤ 6} for
some integer m > 1 we deduce that there exists j, 1 ≤ j ≤ 6 such that there are infinitely
many pairs of primes p, q for which (p − b)/(q − b) = mj (for example p − 1 = 2j (q − 1)
for infinitely many pairs of primes p, q where j = 1, 2, 3, 4, 5 or 6.). There are, presumably,
many other such consequences of Proposition 18.2 to be found.
The main result of this chapter is the following:
Theorem 18.3. For any fixed ² > 0 there are infinitely many pairs of consecutive primes
pn < pn+1 for which pn+1 − pn < ² log pn . In other words
lim inf
n→∞
pn+1 − pn
= 0.
log pn
18.2. Proof of the main results. Define
1
ΛR (n; H, `) =
(k + `)!
X
d|PH (n)
d≤R
µ
¶k+`
R
µ(d) log
.
d
Our results rest on the following estimates:
Proposition 18.4. Suppose H is admissible with |H| = k, and a0 x + b0 ∈ H; and let
h := maxi {|ai |, |bi |}. For R ≤ N 1/2 /(log N )2k and h ≤ RO(1) with R, N → ∞, and any
integer ` ≥ 0, we have
X
n≤N
µ ¶
2` (log R)k+2`
ΛR (n; H, `) =
(S(H) + o(1))N.
`
(k + 2`)!
2
If (18.1) holds with Q = R2 , and h ≤ R² with R, N → ∞, and for any integer ` ≥ 0, we
have
µ
¶
X
2` + 2 (log R)k+2`+1
2
(S(H) + o(1))N.
ΛR (n; H, `) ϑ(a0 n + b0 ) =
` + 1 (k + 2` + 1)!
n≤N
PRIMES
3
Note that if a0 n+b0 is a prime then ΛR (n; H, `) = ΛR (n; H∪{a0 x+b0 }, `) by definition.
We will indicate the proof of this in section 18.5, after showing how such a result can
be applied. We will work out many of the details of the proof (though we will not go into
the details of how the theorem holds uniformly in h as claimed).
18.2. Deduction of the main results from the key proposition. By Proposition
18.4 for ` ≥ 0 and R = N ϑ/2−²/4 , we have
S:=
2N
X
n=N +1
à k
X
!
ϑ(ai n + bi ) − log 3N
ΛR (n; H, `)2
i=1
µ ¶
µ
¶
2` (log R)k+2`
2` + 2 (log R)k+2`+1
(S(H) + o(1))N − log 3N
(S(H) + o(1))N
=k
`
(k + 2`)!
` + 1 (k + 2` + 1)!
µ
¶
µ ¶
2k
2` + 1
(log R)k+2` 2`
=
log R − {1 + o(1)} log 3N
S(H)N
k + 2` + 1 ` + 1
(k + 2`)!
`
µ ¶
µ
¶
` + 1/2
k
(log R)k+2` 2`
≥
· 2ϑ − 1 − ²
S(H)N log 3N.
k + 2` + 1 ` + 1
(k + 2`)!
`
The term inside the brackets is greater than a positive constant (for ² sufficiently small)
`+1/2
1
k
provided k+2`+1
`+1 · 2ϑ > 1. Evidently one can choose such k and ` for any ϑ > 2 ; in
particular one can take ` = 1 and k = 7 when ϑ > 20/21. Thus we deduce Proposition
18.2.
In fact, by the Cauchy-Schwarz inequality, one can show that the number of such n ≤ N
is À N/(log N )C for some constant C.
Exercises
18.2a. Suppose that the Elliott-Halberstam conjecture holds for some ϑ > 1/2. Show that there exists an
integer k, not much larger than 1/(2θ − 1), ˛such that for any admissible k-tuple H, there are infinitely
many n for which at least two members of H˛x=n are prime.
The Bombieri-Vinogradov theorem allows us to take any θ ≤ 1/2 in the above argument
so, after exercise 18.2a, we see that we just fail to prove unconditionally that there are
bounded gaps between consecutive primes. The question therefore becomes as to how we
can “win an ²”. The idea of Goldston, Pintz and Yildirim is to look for prime values
amongst forms a0 x + b0 that do not belong to H. These each contribute much less to
S (by a factor of c log N ) than the forms in H but one wins that needed ² by taking
enough of them. So now let F be a larger set of K forms ai x + bi , containing H, with
h = max1≤i≤K {|ai |, |bi |}.
Using the second part of the key proposition with H replaced by H ∪ {ai x + bi }, as well
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MAT6684
as k by k + 1 and ` by ` − 1 (as discussed just after the statement) we obtain
!
ÃK
2N
X
X
ϑ(ai n + bi ) − log 3N ΛR (n; H, `)2 ,
S̃H :=
i=1
n=N +1
=S+
K
X
2N
X
ΛR (n; H, `)2 ϑ(ai n + bi )
i=k+1 n=N +1
K µ ¶
X
2` (log R)k+2`
=S+
(S(H ∪ {ai x + bi }) + o(1))N
`
(k + 2`)!
i=k+1
√
Therefore taking θ = 1/2 and ` = [ k/2] with k > 5²−2 we obtain that
Áµ
µ ¶ ¶
K
X
(log R)k+2` 2`
S̃H
N ≥
S(H ∪ {ai x + bi }) − 2²S(H) log 3N
(k + 2`)!
`
i=k+1
Now define Sk (F) to be the mean value of S(H) taken over all k-term subsets H of F.
Then the average of the last equation over all such H is
(K − k)Sk+1 (F) − 2²Sk (F) log 3N.
Now when ai x + bi = x + i for each i, Gallagher proved that Sk (F) ∼ 1 as K → ∞, for
each fixed k (see Lemma 18.9 below). Therefore the last displayed equation is > ² log 3N
for K = [4² log 3N ]. Thus there exists n, N < n ≤ 2N such that there are at least two
primes amongst {n + i : 1 ≤ i ≤ 4² log 3N } (once N is sufficiently large), and the result
follows letting ² → 0; that is we have proved Theorem 18.3.
18.3. Lemmas.
Lemma 18.6. (a) We have
X
N <n≤2N
d|PH (n)
µ
1 = νH (d)
¶
N
+ O(1) .
d
(b) Fix A > 0. We also have, for d ≤ (log N )A ,
µ
¶
X
N
N
∗
ϑ(n + h0 ) = νH (d)
+O
,
φ(d)
(log N )A
N <n≤2N
d|PH (n)
∗
∗
where νH
(p) = νH (p) − 1 for each prime p, and νH
(d) is multiplicative in d.
Proof. To see (a) note that for each m (mod d) with d|PH (m) we get N/d + O(1) values
of n ≡ m (mod d) with N < n ≤ 2N (and these are the values of n for which d|PH (n)).
The number of such m is νH (d).
For (b), if we take this same sum where 1 < d < N < n with the additional condition
that n + h0 is prime with h0 ∈ H then d cannot divide n + h0 ; so for each prime p the
∗
number of possibilities for m (mod d) is νH
(d). The result then follows from the SiegelWalfisz theorem.
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5
Lemma 18.7. (Variant of Perron’s formula) For any fixed c > 0 we have
½
Z
0,
if 0 < x ≤ 1
1
xs
(18.2)
ds
=
1
k
2πi (c) sk+1
if x ≥ 1,
k! (log x)
Proof. Exercise 18.3a
Lemma 18.8. Suppose that G(s1 , s2 ) is a double Dirichlet series which is absolutely convergent when Re(s1 ), Re(s2 ) > −1/2, and suppose also that G(0, 0) 6= 0. Then
µ ¶
Z Z
ζ(1 + s1 + s2 )k
Rs1 +s2
(log R)k+2`
1
2`
.
G(s
,
s
)
ds
ds
∼
G(0,
0)
1
2
1
2
(2πi)2 (1) (1)
ζ(1 + s1 )k ζ(1 + s2 )k (s1 s2 )k+`+1
(k + 2`)!
`
1
Sketch of Proof. Let L denote the contour given by s = − 100 log(|t|+2))
+ it: by section
9.6 we know that there are no zeros of ζ(1 + s) on or to the right of L and so that there
are good bounds for ζ, 1/ζ and ζ 0 /ζ in this region. The (difficult) exercise is to bound the
contribution of all of the contours to the error terms in this proof.
To compute the above integral write w = s1 +s2 and s = s1 and let F (s, w) = G(s, w−s),
so that we have
Z
Z
1
ζ(1 + w)k
Rw
F
(s,
w)
dsdw.
(2πi)2 w=(2) s=(1)
ζ(1 + s)k ζ(1 + w − s)k (s(w − s))k+`+1
We move the s-contour to the left to L. The only residue is the pole at s = 0 which
contributes
µ
¶¯
Z
¯
1
1 δ`
H(s, w)
¯ ζ(1 + w)k Rw dw,
¯
`
`+1
2πi w=(2) `! δs
(w − s)
s=0
writing H(s, w) = F (s, w)/((sζ(1 + s))((w − s)ζ(1 + w − s)))k which is analytic in this
domain, and equals
¯
Z
X̀ µ2` − i¶ H(s, w)(i) ¯
1
(wζ(1 + w))k w
¯
=
R dw.
¯
k+2`−i+1
2πi w=(2) i=0
`
i!
s=0 w
We now move the contour over w to the left to L, so that the main term comes from
the unique pole at w = 0, and contributes
µ
¶
X
2` − i H(0, 0)(i,j) am (log R)n
`
i!j!
m!
n!
i+j+m+n=k+2`
i≤`
¯
¡ ∂ ¢i ¡ ∂ ¢j
¯
where H(0, 0)(i,j) = ∂s
H(s,
w)
, and we have written (wζ(1 + w))k =
∂w
s=0,w=0
P
i
i≥0 ai w . The largest power of log R comes from the term with i = 0 and thus we get a
main term of
µ ¶
µ ¶
¯
2`
(log R)k+2`
2`
(log R)k+2`
k¯
H(0, 0)
(wζ(1 + w)) w=0 =
G(0, 0)
`
(k + 2`)!
`
(k + 2`)!
since limw→0 wζ(1 + w) = 1.
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Lemma 18.9. (Gallagher) For any integer k ≤
log log N we have
µ
1
X
k
H⊂{1,...,N }
|H|=k
¡N ¢
√
S(H) = 1 + O
k2
log log N
¶
.
Proof. Let y = c log N for some constant c < 1, and define
¶−k µ
¶
Y µ
1
νH (p)
Sy (H) =
1−
1−
,
p
p
p≤y
so that
¶−k µ
¶µ
¶
Y µ
1
k
k − νH (p)
S(H)/Sy (H) =
1−
1−
1+
.
p
p
p
−
k
p>y
P
P P
Therefore log(S(H)/Sy (H) ¿ p>y k 2 /p2 + i<j p|bj −bi ,p>y p1 . Now 0 < |bj −bi | ≤ N 2
so that there are ≤ 2 log N/ log y such primes p for each pair i and j. Thus, in total, we
k2
k2
2 log N
have ¿ y log
y + k y log y ¿ log log N .
Q
Let m = p≤y p which is = N c+o(1) by the prime number theorem; the value of Sy (H)
depends only upon the value of the elements of H (mod m). Therefore
X
k!
X
Sy (H) =
H⊂{1,...,N }
|H|=k
h1 ,...,hk ∈{1,...,N }
µ
=
X
Sy (H) −
¶k
N
+ O(1)
m
Sy (H)
h1 ,...,hk ∈{1,...,N }
hi =hj for some i6=j
X
Sy (H) + O(k 2 N k−1 (2 log y)k )
h1 ,...,hk ∈{1,...,m}
since each Sy (H) ¿ (2 log y)k . By the Chinese Remainder Theorem we have
1
mk
X
h1 ,...,hk ∈{1,...,N }

Y 1
Sy (H) =
 pk
p≤y
X
h1 ,...,hk ∈{1,...,p}

µ
¶−k µ
¶
1
νH (p)
1−
.
1−

p
p
Now (for the sum over all possible h1 , . . . , hk ∈ {1, . . . , p})
X
(p − νH (p)) =
h
X
X
h
a (mod p)
a6=hi for all i
=
X
X
a (mod p)
h
hi 6=a for all i
Combining the above estimates yields the result.
Exercises
18.3a. Prove Lemma 18.7.
1 = p(p − 1)k .
PRIMES
7
18.3b. Show that the contributions of the integrands on the final contours in Lemma 18.8 are indeed
acceptable.
18.3c. Do the “combining” at the end of the proof of Lemma 18.9
18.4. Proof of Proposition 18.4. We will write “≈” to indicate an error term that
will be considered a little later. In the first part we have
2N
X
¶k+` µ
¶k+`
µ
X
R
1
R
ΛR (n; H, `) =
log
µ(d)µ(e) log
(k + `)!2
d
e
X
2
n=N +1
d,e≤R
≈
=
=
=
1
N <n≤2N
[d,e]|PH (n)
µ
¶k+` µ
¶k+`
X
R
R
1
N
log
log
µ(d)µ(e)ν[d,e] (H)
(k + `)!2
[d, e]
d
e
d,e≤R
Z Z
X µ(d) µ(e) ν[d,e] (H) Rs1
N
R s2
ds1 ds2
(2πi)2 (1) (1)
ds1 es2 [d, e] sk+`+1
s2k+`+1
1
d,e≥1
µ
¶¶
Z Z Yµ
N
νp (H)
1
1
1
Rs1 +s2
1
−
+
−
ds1 ds2
(2πi)2 (1) (1) p
p
ps1
ps2
ps1 +s2
(s1 s2 )k+`+1
Z Z
N
ζ(1 + s1 + s2 )k
Rs1 +s2
G(s
,
s
)
ds1 ds2
1
2
(2πi)2 (1) (1)
ζ(1 + s1 )k ζ(1 + s2 )k (s1 s2 )k+`+1
where
G(s1 , s2 ) =
³
Y 1−
p
νp (H)
p
´´ ³
1−
− ps11+s2
´k ³
´k
³
1
1
1 − p1+s
1
−
1+s
1
2
p
³
1
ps1
+
1
p s2
1
p1+s1 +s2
´k
,
which is absolutely convergent when Re(s1 ), Re(s2 ) > −1/2. The result follows from
lemma 18.8 since G(0, 0) = S(H), after we have justified the “≈”: By lemma 18.6(a), the
error term here is
(log R)2(k+`) X
(log R)2(k+`) X
RR
2
2
µ(d)
µ(e)
ν
(H)
≤
µ(d)2 µ(e)2 ν[d,e] (H)
[d,e]
2
2
(k + `)!
(k + `)!
d e
d,e≤R
d,e≤R
µ
µ
¶¶
¶
2(k+`) Y
Yµ
2
1
2k
2 (log R)
2
2(k+`)
≤R
1 + νp (H)
+
¿k R (log R)
1+
(k + `)!2
p p2
p
¿
p≤R
2
4k+2`
¿ R (log R)
and this is negligible provided R ≤ N 1/2 /(log N )2k .
p≤R
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MAT6684
For the second part we have, proceeding analogously to the above,
2N
X
n=N +1
ΛR (n; H, `)2 ϑ(n + h0 )
¶k+` µ
¶k+`
µ
X
1
R
R
=
µ(d)µ(e) log
log
(k + `)!2
d
e
d,e≤R
X
θ(n + h0 )
N <n≤2N
[d,e]|PH (n)
µ
¶k+` µ
¶k+`
X
1
N
R
R
∗
≈
log
log
µ(d)µ(e)ν[d,e] (H)
(k + `)!2
φ([d, e])
d
e
d,e≤R
µ
¶¶
Z Z Yµ
N
(νp (H) − 1)
1
1
1
Rs1 +s2
1
−
ds1 ds2
=
+
−
(2πi)2 (1) (1) p
(p − 1)
ps1
ps2
ps1 +s2
(s1 s2 )k+`+1
Z Z
N
Rs1 +s2
ζ(1 + s1 + s2 )k−1
∗
=
G
(s
,
s
)
ds1 ds2
1
2
(2πi)2 (1) (1)
ζ(1 + s1 )k−1 ζ(1 + s2 )k−1 (s1 s2 )k+`+1
where
³
G∗ (s1 , s2 ) =
Y 1−
(νp (H)−1)
(p−1)
³
p
³
1
p s1
+
1
p s2
1
´´ ³
1
− ps1 +s2
1 − p1+s1 +s2
´k−1 ³
´k−1
1
1
1 − p1+s
1 − p1+s
1
2
´k−1
,
which is absolutely convergent when Re(s1 ), Re(s2 ) > −1/2. The result follows from
lemma 18.8 (with k − 1 in place of k, and ` + 1 in place of `) since G∗ (0, 0) = S(H), after
we have justified the “≈”: By lemma 18.6(b), the error term here is
¯
¯
¯
¯
(log R)2(k+`) X
N
2
2 ∗
¯θ(N ; [d, e], a) −
¯
≤
µ(d)
µ(e)
ν
(H)
max
[d,e]
(k + `)!2
φ([d, e]) ¯
a:(a,[d,e])=1 ¯
d,e≤R
¯
¯
¯
N ¯¯
(log R)2(k+`) X
2
∗
¯
µ(m) τ (m)νm (H) max ¯θ(N ; m, a) −
≤
(k + `)!2
φ(m) ¯
a:(a,m)=1
2
m≤R
The square of this sum is, using Cauchy-Schwartz and the trivial upper bound θ(N ; m, a) ¿
(N/m) log N ,
¯
¯2
¯
N ¯¯
¯
≤
µ(m)
·
max m ¯θ(N ; m, a) −
m
φ(m) ¯
a:(a,m)=1
2
2
m≤R
m≤R
!
Ã
¯
¯
X
Y
¯
τ (p)2 νp∗ (H)2
N ¯¯
¯
·
max θ(N ; m, a) −
¿
1+
N log N
p
φ(m) ¯
a:(a,m)=1 ¯
2
2
m≤R
p≤R
¯
¯
X
¯
¯
2
N
¯.
max ¯¯θ(N ; m, a) −
¿ (log R)4(k−1) N log N
¯
φ(m)
a:(a,m)=1
2
X
2 τ (m)
2 ∗
νm (H)2
m≤R
X
PRIMES
9
Inserting this above we get that our error term is negligible provided
¯
¯
X
¯
N ¯¯
N
¯
max ¯θ(N ; m, a) −
¿
¯
φ(m)
(log N )A
a:(a,m)=1
2
m≤R
for some A > 4k 2 − 6k + 3. This holds by the Bombieri–Vinogradov theorem for R ≤
N 1/4 /(log N )B(k) ; and by the Elliott-Halberstam conjecture for R ≤ N 1/2−² .